Question on Dirac Delta and functional derivatives

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Main Question or Discussion Point

Suppose that we have a compact manifold [itex]\mathcal{M}[/itex] with a positive definite metric [itex]g_{ij}[/itex]. The volume of the manifold is then

[tex]
V = \int_\mathcal{M} d^3x \sqrt{g(x)},
[/tex]

where [itex]x^i[/itex] are coordinates on [itex]\mathcal{M}[/itex] and [itex]\sqrt{g(x)}[/itex] is the square root of the determinant of the metric. Suppose now that we take a functional derivative of the volume with respect to the metric, i.e., we want to calculate

[tex]
\frac{\delta}{\delta g_{ij}(y)} V
= \frac{\delta}{\delta g_{ij}(y)}\int_\mathcal{M} d^3x \sqrt{g(x)}
[/tex]

As far as I know, this calculation gives

[tex]
\frac{\delta}{\delta g_{ij}(y)} V
= \frac{1}{2}\int_\mathcal{M} d^3x \sqrt{g(x)} g^{ij}(x) \delta^{(3)}(x,y)
[/tex]

Where [itex]\delta^{(3)}(x)[/itex] is a three-dimensional Dirac delta function. The question I have is this: is [itex]\delta^{(3)}(x,y)[/itex] to be regarded as (a) a unit weight density or (b) a density of weight zero?

If the answer is (a) then the result is

[tex]
\frac{\delta}{\delta g_{ij}(y)} V
= \frac{1}{2}\sqrt{g(y)} g^{ij}(y)
[/tex]

while if the answer is (b) then the result for the functional derivative is

[tex]
\frac{\delta}{\delta g_{ij}(y)} V
= \frac{1}{2} g^{ij}(y)
[/tex]

So which one is the correct answer?
 
Last edited:

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